Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z + 8}{-2z + 6} \times \dfrac{z^2 - 12z + 27}{3z^2 - 39z + 108} $
First factor out any common factors. $x = \dfrac{z + 8}{-2(z - 3)} \times \dfrac{z^2 - 12z + 27}{3(z^2 - 13z + 36)} $ Then factor the quadratic expressions. $x = \dfrac {z + 8} {-2(z - 3)} \times \dfrac {(z - 9)(z - 3)} {3(z - 9)(z - 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {(z + 8) \times (z - 9)(z - 3) } {-2(z - 3) \times 3(z - 9)(z - 4) } $ $x = \dfrac {(z - 9)(z - 3)(z + 8)} {-6(z - 9)(z - 4)(z - 3)} $ Notice that $(z - 9)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {\cancel{(z - 9)}(z - 3)(z + 8)} {-6\cancel{(z - 9)}(z - 4)(z - 3)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $x = \dfrac {\cancel{(z - 9)}\cancel{(z - 3)}(z + 8)} {-6\cancel{(z - 9)}(z - 4)\cancel{(z - 3)}} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $x = \dfrac {z + 8} {-6(z - 4)} $ $ x = \dfrac{-(z + 8)}{6(z - 4)}; z \neq 9; z \neq 3 $